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Calculate the minimum bit rate for a channel having b.w. 3100 Hz and S/N ratio 10 dB
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Given: - B = 3100 Hz

$\big(\frac{S}{N}\big)dB=10$

$∴10=10 log_{10}⁡big(\frac{S}{N}\big) \\ ∴\frac{S}{N}=10$

∴ Maximum bit rate

$= R_{\max} = B \log_2⁡ \big[1+\frac{S}{N}\big] \\ = 3100 \log_2⁡(1+10) \\ =\frac{3100 \log_{10}⁡11}{\log_{10}⁡2} \\ = 10,724 bits/sec$

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